2019-2020学年新一线同步人教A版数学必修一练习：5.5.1　第2课时　两角和与差的正弦、余弦、正切公式 含解析.docx

教学资料范本2019-2020学年新一线同步人教A版数学必修一练习：5.5.1第2课时两角和与差的正弦、余弦、正切公式 含解析编 辑：_时 间：_2课时两角和与差的正弦、余弦、正切公式课后篇巩固提升基础巩固1.sin 35cos 5-cos 35sin 5=()A.B.1C.2D.2sin 40解析sin 35cos 5-cos 35sin 5=sin(35-5)=sin 30=.答案A2.若sin=cos,则tan =()A.-1B.0C.D.1解析由已知得cos -sin =cos -sin ,因此sin =cos ,于是tan =-1.答案A3.若tan(+)=,tan(-)=,则tan 2=()A.B.C.D.解析tan 2=tan (+)+(-)=.答案D4.sin(+75)+cos(+45)-cos(+15)的值等于()A.1B.1C.-1D.0解析原式=sin (+45)+30+cos(+45)-cos (+45)-30=sin(+45)+cos(+45)+cos(+45)-=sin(+45)+cos(+45)-cos(+45)-sin(+45)=0.答案D5.设,且tan =,则()A.3-=B.3+=C.2-=D.2+=解析由tan =,得,得sin cos -cos sin =cos ,sin(-)=sin.又,故-=-,即2-=.答案C6.化简:=.解析原式=-1.答案-17.(一题多空题)已知锐角,满足(tan -1)(tan -1)=2,则tan(+)=,+=.解析因为(tan -1)(tan -1)=2,所以tan +tan =tan tan -1.因此tan(+)=-1,因为+(0,),所以+=.答案-18.已知,tan =2,则cos=.解析由tan =2,得sin =2cos .又sin2+cos2=1,cos =,sin =.cos=cos cos+sin sin=.答案9.tan 23+tan 37+tan 23tan 37的值是.解析tan 60=,tan 23+tan 37=tan 23tan 37,tan 23+tan 37+tan 23tan 37=.答案10.化简求值:(1)sin(+)cos(-)+cos(+)sin(-);(2)cos(70+)sin(170-)-sin(70+)cos(10+);(3)cos 21cos 24+sin 159sin 204.解(1)原式=sin(+-)=sin 2.(2)原式=cos(70+)sin(10+)-sin(70+)cos(10+)=sin (10+)-(70+)=sin(-60)=-.(3)原式=cos 21cos 24+sin(180-21)sin(180+24)=cos 21cos 24-sin 21sin 24=cos(21+24)=cos 45=.11.已知,均为锐角,且tan =,求tan(+)的值.解tan =tan,因为,均为锐角,所以-,0,又y=tan x在上是单调函数,所以=-,即+=,tan(+)=1.能力提升1.已知,tan=-3,则sin =()A.B.-C.D.解析tan =tan =-,因为,所以,故sin =.答案A2.设,都为锐角,且cos =,sin(+)=,则sin 等于()A.B.C.D.-解析为锐角,cos =,sin =.,都为锐角,0+.sin(+)=,cos(+)=.当cos(+)=-时,sin =sin(+)-=sin(+)cos -cos(+)sin =;当cos(+)=时,sin =sin(+)-=sin(+)cos -cos(+)sin =-,与已知为锐角矛盾.sin =.答案B3.已知cos(+)=,cos(-)=-,则cos cos =.解析由已知得cos cos -sin sin =,cos cos +sin sin =-,两式相加得2cos cos =0,故cos cos =0.答案04.已